A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 Kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post’s length when the load is applied?

Short Answer

Expert verified

(a)-1.60×106Pa(b)-8×10-6(c)-2.0×10-5m

Step by step solution

01

Given information

Length I0= 2.50m, diameter d = 2.5x10-2m, Load m = 8000kg

02

Concept/Formula used

Y=l0FAΔl

Where, Y is Young’s modulus, I0 is length of muscle, F is muscle force, A is cross-sectional area andl is elongation.

03

Cross –sectional area of the post

A=πd24=π2.50×10-2m24=0.05m2

04

Force application on the post

F=mg=8000kg9.80m/s2=7.84×104NYoungsModulusforsteelisY=2.0×1011Pa

05

Stress in the post

(a)

stress=FA=-7.84×104N0.05m2=-1.60×106Pa

Note: negative sign shows that the stress is compressive.

06

Strain in the post

(b)

strain=stressY=-1.60×106Pa2×1011Pa=-8×10-6

Note: negative sign shows that the length decreases.

07

Change in the post length

(C)

Δl=l0strain=2.5m-8×10-6=-2.0×10-5m

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